Let depth of water in cylinder be $x$
So velocity $(v)$ of efflux $=\sqrt{2 g x}$
Time taken $(t)$ by water to travel vertical distance of $H$
$=\sqrt{\frac{2 H}{g}}$
$\Rightarrow \text { Range }=v \times t$
$R=\sqrt{2 g x} \times \sqrt{\frac{2 H}{g}}$
Solving this we get
$\frac{R^2}{4 H}=x$



